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3 years ago

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A new branch in its first year opened new accounts at a rate of 40 for every 100 customers calls. The total number of new accoun

ts opened in the year was 10,000. How many customers calls were made?

1 answer:

Mumz [18]3 years ago

4 0

40 new accounts for: 100 customer calls
10,000 new accounts for: x customer calls.
----------------------------------------------------------------
40 : 10,000 = 100 : x
x = (100 · 10,000 ) : 40 = 25,000
Answer: 25,000 customer calls were made.

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Find the circumference of a circle with radius, r = 19.5m.<br>Give your answer in terms of π.

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Answer:

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Step-by-step explanation:

The circumference of a circle is given by

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C = 2 * pi * 19.5

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A Euler circuit has ___ ODD vertices

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Answer:

A Euler circuit has TWO ODD vertices

Step-by-step explanation:

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second option

G(3 , -4) ; H(8 , 1)

Step-by-step explanation:

For the line GH to be a median, the point G must have the intermediate value at x and at y between points A and B, and the point H must have the intermediate value at x and at y between points A and B

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3 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

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Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

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Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

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3 years ago